如何使用parsec解析Haskell中的Tuple（String，Int） [英] How to parse a Tuple (String,Int) in Haskell using parsec

问题描述

我想我已经设法将字符串和字符串解析为Ints，但是我还需要解析（String，Int）类型，因为userRatings是为了正确读取textFile并且我正在使用Parsec

这是解析以及导入

  import Text.Parsec 
（Parsec，ParseError，parse  - 类型和解析器
，between，noneOf，sepBy，many1  - 组合器
，char，空格，数字，换行符 - 简单解析器
）
 
  - 将字符串解析为字符串
 stringLit :: Parsec字符串u字符串
 stringLit =之间（char'''）（char'''）$ many1 $ noneOf\\ \\n
 
  - 将字符串解析为字符串列表
 listOfStrings :: Parsec字符串u [String] 
 listOfStrings = stringLit`sepBy`（char' ，>>空格）
 
  - 将字符串解析为一个int 
 intLit :: Parsec字符串u Int 
 intLit = fmap读取$ many1数字
 - 或`用Control.Applic读取< $> many1 digit` ative 
 
 film :: Parsec String u电影
电影=做
  - 或者`title<  -  stringLit< * newline` with Control.Applicative 
 title <  -  stringLit 
 newline 
 director<  -  stringLit 
 newline 
 year<  -  intLit 
 newline 
 userRatings<  -  listOfStrings 
 newline 
 return（title，director，year，userRatings）
  

解决方案

您可以通过使用 Applicative 或 Monad 接口从现有解析器进行组合来完成此操作。 解析器具有两者的实例。

使用应用： $ b $>

  stringIntTuple :: Parser（String，Int）
 stringIntTuple =（，）< $> yourStringParser< *> yourIntParser 
  

这与以下内容相同：

  stringIntTuple :: Parser（String，Int）
 stringIntTuple = liftA2（，）yourStringParser yourIntParser 
  

使用 Monad ：

  stringIntTuple :: Parser（String，Int）
 stringIntTuple = 
 do 
 theString<  -  yourStringParser 
 theInt<  -  yourIntParser 
 return（theString，theInt） 
  

I think i already managed to parse strings to strings and strings to Ints but I also need to parse a (String,Int) type as userRatings is in order to read from a textFile correctly and I am using Parsec

This is the Parsing along with the import

import Text.Parsec
( Parsec, ParseError, parse        -- Types and parser
, between, noneOf, sepBy, many1    -- Combinators
, char, spaces, digit, newline     -- Simple parsers
)

-- Parse a string to a string
stringLit :: Parsec String u String
stringLit = between (char '"') (char '"') $ many1 $ noneOf "\"\n"

-- Parse a string to a list of strings
listOfStrings :: Parsec String u [String]
listOfStrings = stringLit `sepBy` (char ',' >> spaces)

-- Parse a string to an int
intLit :: Parsec String u Int
intLit = fmap read $ many1 digit
-- Or `read <$> many1 digit` with Control.Applicative

film :: Parsec String u Film
film = do
-- alternatively `title <- stringLit <* newline` with Control.Applicative
title <- stringLit
newline
director <- stringLit
newline
year <- intLit
newline
userRatings <- listOfStrings
newline
return (title, director, year, userRatings)

解决方案

You can do this by composing from existing parsers using either the Applicative or the Monad interface. Parser has instances of both.

Using Applicative:

stringIntTuple :: Parser (String, Int)
stringIntTuple = (,) <$> yourStringParser <*> yourIntParser

Which is the same as the following:

stringIntTuple :: Parser (String, Int)
stringIntTuple = liftA2 (,) yourStringParser yourIntParser

Using Monad:

stringIntTuple :: Parser (String, Int)
stringIntTuple =
  do
    theString <- yourStringParser
    theInt <- yourIntParser
    return (theString, theInt)

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